# a lot contains 20 articles.d probability that d lot  contains xactly 2 defective articles is 0.4;and d probability dat it contains exactly  3 defective articles is 0.6.articles r drawn from d lot at random 1 by 1 without replacement until all d defective r found.Wat is d probability dat d testing procedure stops at d twelfth testing?

Yash Baheti IIT Roorkee
8 years ago
Hi,

Clearly there are two cases, either the lot as 2 deffective articles or 3 as the sum of probabilities is 1.

Now case 1 : When the lot has 2 defective articles. So we are going to draw one defective article between deaw no 1 to 11th and will draw the second defective article at 12th draw as this will mark the end of our search for this case.

So the probability will be 18/20.17/19 …...9/11.(2/10).1/9 (total 12 draws in which we have taken perfect articles till 10th draw, on 11th we took one defective and on twelth we took another defective)

Now we need to arrange one defective draw in 10 perfect article draws which will be : (11!)/(10!.1!)

Multiply this with the above series, this is result for first case,

Similary for the second case we have : 17/20.16/19.......9/12 . 3/11 . 2/10. 1/9 (i.e. 9 perfect article draws and there on 3 defective draws, now we need to keep the twelth draw at its place and have to permute rest 11 draws in which 9 are perfect draws and 2 are defective so : (11!)/(9!.2!) Multiply this with the second case series.

Add the results of two cases, which will be ur answer. ie:
18/20.17/19 …...9/11.(2/10).1/9[(11!)/(10!.1!)] + 17/20.16/19.......9/12 . 3/11 . 2/10. 1/9[ (11!)/(9!.2!)]